Subgraph Complementation Fedor V

Subgraph Complementation Fedor V. Fomin, Petr A. Golovach, Torstein Strømme, Dimitrios M. Thilikos

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Fedor V. Fomin, Petr A. Golovach, Torstein Strømme, Dimitrios M. Thilikos. Subgraph Comple- mentation. Algorithmica, Springer Verlag, 2020, 82 (7), pp.1859-1880. 10.1007/s00453-020-00677-8. hal-03002656

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Distributed under a Creative Commons Attribution| 4.0 International License Algorithmica https://doi.org/10.1007/s00453-020-00677-8

Subgraph Complementation

Fedor V. Fomin1 · Petr A. Golovach1 · Torstein J. F. Strømme1 Dimitrios M. Thilikos2

Received: 30 January 2019 / Accepted: 8 January 2020 © The Author(s) 2020

Abstract A subgraph complement of the graph G is a graph obtained from G by complementing all the edges in one of its induced subgraphs. We study the following algorithmic question: for a given graph G and graph class G , is there a subgraph complement of G which is in G ? We show that this problem can be solved in polynomial time for various choices of the graphs class G , such as bipartite, d-degenerate, or cographs. We complement these results by proving that the problem is NP-complete when G is the class of regular graphs.

Keywords Subgraph complementation · Graph editing · Graph classes

Mathematics Subject Classifcation 05C85 · 68Q25

1 Introduction

One of the most important questions in graph theory concerns the efciency of recognition of a graph class G . For example, how fast we can decide whether a graph is chordal, 2-connected, triangle-free, of bounded treewidth, bipartite, 3-colorable, or excludes some fxed graph as a minor? In particular, the recent developments in parameterized algorithms are partially driven by the problems of recognizing of graph classes which difer only up to a “small disturbance” from graph

* Torstein J. F. Strømme [email protected] Fedor V. Fomin [email protected] Petr A. Golovach [email protected] Dimitrios M. Thilikos [email protected]

1 University of Bergen, Bergen, Norway 2 LIRMM, Univ. Montpellier, CNRS, Montpellier, France Algorithmica classes recognizable in polynomial time. The amount of disturbance is quantifed in “atomic” operations required for modifying an input graph into the “well-behaving” graph class G . The standard operations could be edge/vertex deletions, additions or edge contractions. Many problems in graph algorithms fall into this graph modifcation category: is it possible to add at most k edges to make a graph 2-edge connected or to make it chordal? Or is it possible to delete at most k vertices such that the resulting graph has no edges or contains no cycles? A rich subclass of modifcation problems concerns edge editing problems. Here the “atomic” operation is the change of adjacency, i.e. for a pair of vertices u, v, we can either add an edge uv or delete the edge uv. For example, the Cluster Editing problem asks to transform an input graph into a cluster graph—that is, a disjoint union of cliques—by fipping at most k adjacency relations. Besides the basic edge editing, it is natural to consider problems where the set of removed and added edges should satisfy some structural constraints. In particular, such problems were considered for complementation problems. Recall that the complement of a graph G is a graph H on the same vertices such that two distinct vertices of H are adjacent if and only if they are not adjacent in G. Seidel (see [22–24]) introduced the operation that is now known as the Seidel switch. For a vertex v of a graph G, this operation complements the adjacencies of v, that is, it removes the edges incident to v and makes v adjacent to the non-neighbors of v in G. Seidel switching a set of vertices U entails consecutively switching each vertex in the set. The result is that all adjacencies between U and its complement V(G)⧵U are fipped. The study of the algorithmic question whether it is possible to obtain a graph from a given graph class by the Seidel switch was initiated by Ehrenfeucht et al. [7]. Fur- ther results were established in [14–16, 18, 19]. Another important operation of this type is the local complementation. For a vertex v of a graph G, the local complementation of G at v is the graph obtained from G by replacing G[N(v)] by its complement. This operation plays crucial role in the def- inition of vertex-minors [20] and was investigated in this context (see, e.g. [6, 21]). See also [2, 17] for some algorithmic results concerning local complementations. In this paper we study the subgraph complement of a graph, which was introduced by Kamiński et al. [17] in their study of the clique-width of a graph. A subgraph complement of a graph G is a graph obtained from G by complementing all the edges of one of its induced subgraphs. More formally, for a graph G and S ⊆ V(G) , we defne G ⊕ S as the graph with the vertex set V(G) whose edge set is defned as follows: a pair of distinct vertices u, v is an edge of G ⊕ S if and only if one of the following holds:

• uv ∈ E(G)∧(u ∉ S ∨ v ∉ S) , or • uv ∉ E(G)∧u ∈ S ∧ v ∈ S.

Thus when the set S consists only of two vertices {u, v} , the operation simply changes the adjacency between u and v, and for a larger set S, G ⊕ S changes the adjacency relations for all pairs of vertices of S. We say that a graph H is a subgraph complement of the graph G if H is isomorphic to G ⊕ S for some S ⊆ V(G) . For a graph class G and a graph G, we say that Algorithmica there is a subgraph complement of G to G if for some S ⊆ V(G) , we have G ⊕ S ∈ G . (1) We denote by G the class of graphs such that its members can be subgraph complemented to G. Let G be a graph class. We consider the following generic algorithmic problem.

(1) In other words, how difcult is it to recognize the class G ? In this paper we (1) show that there are many well-known graph classes G such that G is recognizable in polynomial time. We show that

G O ⋅ 4 6 G • Subgraph Complement to is solvable in (f (n) n + n ) time when is a triangle-free graph class recognizable in f(n) time. For example, this implies that (1) when G is the class of bipartite graphs, the class G is recognizable in polynomial time. This result is found in Sect. 3. G ⋅ 2O(d) G • Subgraph Complement to is solvable in f (n) n time when is a d-degenerate graph class recognizable in f(n) time. Thus when G is the class of planar graphs, class of cubic graphs, class of graph of bounded treewidth, or class of (1) H-minor free graphs, then the class G is recognizable in polynomial time. This result is found in Sect. 4. G G • Subgraph Complement to is solvable in polynomial time when is a class of bounded clique-width expressible in monadic second-order logic (with no edge G set quantifcation). In particular, if is the class of P4-free graphs (cographs), (1) then G is recognizable in polynomial time. This result is found in Sect. 6. G G • Subgraph Complement to is solvable in polynomial time when can be (1) described by a 2 × 2 M-partition matrix. Therefore G is recognizable in polynomial time when G is the class of split graphs, as they can be described by such a matrix. This result is found in Sect. 5.

There are nevertheless cases when the problem is NP-hard. In particular, we prove that this holds when G is the class of regular graphs. This result is found in Sect. 7.

2 Preliminaries

+ We let ℕ∶={0, 1, 2 …} be the set of the natural numbers, and we let ℕ ∶=ℕ⧵{0} be the set of positive integers. For a set A and a non-negative integer k ∈ ℕ , we denote A A k ∶={A� ⊆ A ∣ A� = k} the family of all subsets of of size as k . Simple Graph A (simple) graph is a pair G∶=(V, E) , where V is a set of vertices V E ⊆ edges {u, v}∈E(G) uv and 2 is a set of . For an edge we use the shorthand Algorithmica

(or equivalently, vu). For a given graph G, we refer to its vertex set as V(G), and its edge set as E(G). For a set of edges F ⊆ E(G) , we denote by V(F) the set of vertices that are contained in the edges of F, i.e. V(F)∶= uv∈F{u, v}. (Induced) Subgraph For graphs G and H we⋃ say that H is a subgraph of G, denoted by H ⊆ G if V(H) ⊆ V(G) and E(H) ⊆ E(G) . For a set of vertices A ⊆ V(G) , A G A G induced A G[A]∶=(A, E(G)∩ ) we denote by [ ] the subgraph of by , i.e. 2 . We use the notation G − X∶=G[V(G)⧵X] and for a set of edges F ⊆ E(G) , G − F∶=(V(G), E(G)⧵F) . For a vertex v ∈ V(G) (or an edge e ∈ E(G) ), we use the shorthand G − x∶=G −{x} (or G − e∶=G −{e} , respectively). Neighborhood Let G be a graph. For a vertex v ∈ V(G) , we denote by NG(v) the open neighborhood of v, i.e. NG(v)∶={w ∣ vw ∈ E(G)} , and by NG[v] the closed neighborhood of v, i.e. NG[v]∶={v}∪NG(v) . The degree of v is the size of its open neighborhood, i.e. degG(v)∶= NG(v) . For a set of vertices X ⊆ V(G) , we let NG(X)∶= v∈X NG(v)⧵X and NG[X]∶=NG(X)∪X . We use the shorthand notations ‘N’ and ‘ deg⋃ ’ for ‘ NG ’ and ‘ degG ’, respectively, if G is clear from the context. Monadic Second Order Logic 1 is a restricted version of of 2 (Monadic Second Order Logic) without quantifcations over edge subsets. More precisely, the ⇔ ⇒ syntax of 1 of graphs includes the logical connectives ∨, ∧, ¬, , , variables for vertices and sets of vertices, the quantifers ∀, ∃ that can be applied to these variables, and the following binary relations:

1. u ∈ U where u is a vertex variable and U is a vertex set variable; 2. (u, v), where u and v are vertex variables and the interpretation is that u and v are adjacent; 3. equality of variables representing vertices and sets of vertices.

We refer to [3] for more information on 1 and 2.

3 Subgraph Complementation to Triangle‑Free Graph Classes

A triangle is a complete graph on three vertices. Many graph classes do not allow the triangle as a subgraph, for instance trees, forests, or graphs with large girth. In this section we show that subgraph complementation to triangle-free graphs can be decided in polynomial time. More precisely, we show that if a graph class G can be recognized in polynomial G time and it is triangle-free, then we can also solve Subgraph Complement to in polynomial time. Our algorithm is constructive, and returns a solution S ⊆ V(G) , that is a set S such that G ⊕ S is in G . We say that a solution hits an edge uv (or a non-edge uv) , if both u and v are contained in S. Our algorithm considers each of the following cases.

(i) There is a solution S containing two vertices that are non-adjacent in G. (ii) There is a solution S such that it forms a clique of size at least 2 in G. (iii) G is a no-instance. Algorithmica

We start from analyzing the structure of a solution in Case (i). We need the following observation.

Observation 1 Let G be a class of triangle-free graphs and let G be an instance of G Subgraph Complement to , where S ⊆ V(G) is a valid solution. Then

(a) G[S] does not contain an independent set of size 3, and (b) for every triangle {u, v, w} ⊆ V(G) , at least two vertices are in S.

Because all non-edges between vertices in G[S] become edges in G ⊕ S and vice versa, whereas all (non-) edges with an endpoint outside S remain untouched, we see that the observation holds. Before delving into the analysis of Case (i) any further, let us recall that a graph G is a split graph if its vertex set can be partitioned into V(G)=C ∪ I , where C is a clique and I is an independent set. Let us note that the vertex set of a split graph can have several split partitions, i.e. partitions into a clique and independent set. However, the number of split partitions of an n-vertex split graphs is at most n + 1 , and they can be enumerated in linear time. This builds on the result of Hammer and Simone [12] (see also [11]), for completeness we provide a proof.

Lemma 2 Let G be a split graph on n vertices. Then G has at most n + 1 split partitions. Moreover, they can be enumerated in linear time.

Proof Let denote the cardinality of a maximum clique in G. It is known that every split partition has a clique of size or − 1 [12]. We will show that there are at most n partitions where the clique has cardinality , and that these can be enumerated in linear time—since the split partitions in G and its complement G are the same up to swapping the clique and independent set, this is enough to prove the lemma with a bound of 2n. We will strengthen the bound at the end of the proof. Let V≥ ⊆ V(G) denote the set of vertices whose degree is at least . We observe that in every split partition, these vertices need to be in the clique; otherwise there is either an edge between two vertices of the independent set, or it is possible to fnd a clique of size + 1. Similarly, let V≤ −2 ⊆ V(G) denote the set of vertices whose degree is at most − 2 . We observe that every vertex of V≤ −2 must be in the independent set of a split partition where the clique has size . It remains to partition the set of vertices V −1 ⊆ V(G) whose degrees are − 1 . We claim that every vertex of V −1 must have V≥ contained in its neighborhood, or else there is a unique split partition whose clique of size is exactly V≥ . Assume that there is a vertex u ∈ V −1 whose neighborhood does not include a vertex v ∈ V≥ . Since v is in every clique of size , u can never be in the clique, implying that all of u’s neighbors must be—but then those neighbors have degree at least , implying that V≥ ≥ . We proceed with the case when V≥ < . Consider some subset U ⊆ V −1 such that U ∪ V≥ is a clique of size . Then U is a clique, and vertices of U have no Algorithmica

neighbors in V −1⧵U . Thus G[V −1] is a collection of cliques, each of size − V≥ . Each split partition of size will include exactly one of these cliques in addition to V≥ . However, observe that this implies that there is either exactly one clique, or the cliques must have size 1—otherwise there would be an edge between two vertices in the independent set. This shows that there are at most n distinct split partitions whose clique has size . In order to enumerate the partitions in linear time, we frst analyze the degree sequence to fnd using the result of Hammer and Simone [12] (sorting vertices by O degree can be done in (n) time using bucket sort). We then fnd the sets V≥ and V −1 . If V≥ = , we have a unique partition; if V≥ ∪ V −1 = we also have a unique partition; otherwise V≥ = − 1 , and V −1 is an independent set—the split partitions with a clique of size have one pick of vertex from V −1 each. It remains to tighten the bound on the number of split partitions to n + 1 . We show that if there is more than one split partition with a clique of size , then there is a single split partition with a clique of size − 1 . In more detail: if there is more than one split partition with a clique of maximum size, we know from the above par- agraph that V −1 is an independent set of size at least two, and V≥ is a clique of size − 1 . We claim that every clique of size − 1 in a split partition contains V≥ , thus making the split partition unique. Assume for the sake of contradiction that u ∈ V≥ is not in the clique. Recall that u is in the neighborhood of every vertex in V −1 , so a split partition placing u in the independent set in must place V −1 in the clique. But V −1 is an independent set of size at least two, so this is impossible. We remark that the bound is tight, and can be obtained by an edgeless graph. ◻

Returning to the problem of subgraph complement to triangle free graph classes, we are now ready for the analysis of Case (i): when there is a solution S containing two vertices that are non-adjacent in G. G Lemma 3 Let be a class of triangle-free graphs and let G be an instance of Sub- G graph Complement to . Let S ⊆ V(G) be a valid solution which is not a clique, and let u, v ∈ S be distinct vertices such that uv ∉ E(G) . Then

(a) the solution S is a subset of the union of the closed neighborhoods of u and v, that is S ⊆ NG[u]∪NG[v]; (b) every common neighbor of u and v must be contained in the solution S, that is NG(u)∩NG(v) ⊆ S; (c) the graph G[N(u)⧵N(v)] is a split graph. Moreover, (N(u)⧵N(v)) ∩ S is a clique and (N(u)⧵N(v))⧵S is an independent set.

Proof We will prove each point separately, and in order.

(a) Assume for the sake of contradiction that the solution S contains a vertex w ∉ NG[u]∪NG[v] . But then {u, v, w} is an independent set in G, which contradicts Observation 1a. Algorithmica

(b) Assume for the sake of contradiction that the solution S does not contain a vertex w ∈ NG(u)∩NG(v) . Then the edges uw and vw will both be present in G ⊕ S , as well as the edge uv. Together, these form a triangle. (c) We frst claim that the solution S is a vertex cover for G[N(u)⧵N(v)] . If it was not, then there would exist an edge u1u2 of G[N(u)⧵N(v)] such that both endpoints u1, u2 ∉ S , yet u1, u2 would form a triangle with u in G ⊕ S , which would be a contradiction. Hence (N(u)⧵N(v))⧵S is an independent set. Secondly, we claim that (N(u)⧵N(v)) ∩ S forms a clique. If not, then there would exist u1, u2 ∈(N(u)⧵N(v)) ∩ S which are nonadjacent. In this case {u1, u2, v} is an independent set, which contradicts Observation 1a. Taken together, these claims imply the last item of the lemma.

◻ We now move on to examine the structure of a solution for Case (ii), when there exists a solution which is a clique. Note that we can assume the clique has size at least two, since if S ≤ 1 then subgraph complementation does not alter the graph. G Lemma 4 Let be a class of triangle-free graphs and let G be an instance of Sub- G graph Complement to . Let S ⊆ V(G) be a solution such that S ≥ 2 and G[S] is a clique. Let u, v ∈ S be distinct. Then (a) the solution S is contained in their common neighborhood, that is S ⊆ NG[u]∩NG[v] , and (b) the graph G[NG[u]∩NG[v]] is a split graph where (NG[u]∩NG[v])⧵S is an independent set.

Proof We prove each point separately, and in order.

(a) Assume for the sake of contradiction that the solution S contains a vertex w which is not in the neighborhood of both u and v. This contradicts that S is a clique. (b) We claim that S is a vertex cover of G[NG[u]∩NG[v]] . Because S is also a clique, the statement of the lemma will then follow immediately. Assume for the sake of contradiction that S is not a vertex cover. Then there exist an uncovered edge w1w2 , where w1, w2 ∈ NG[u]∩NG[v] , and also w1, w2 ∉ S . Since {u, w1, w2} form a triangle, we have by Observation 1b that at least two of these vertices are in S. That is a contradiction, so our claim holds. ◻

We now have everything in place to present the algorithm. G G Algorithm 5 (Subgraph Complement to where is triangle-free) Algorithmica

Input: An instance G of SC G where G is a triangle-free graph class recognizable in f(n) time for some function f. Output: A set S ⊆ V(G) such that G ⊕ S is in G , or a correct report that no such set exists.

1. For every non-edge uv of G:

(a) If either G[N(u)⧵N(v)] or G[N(u)⧵N(v)] is not a split graph, skip this iteration and try the next non-edge. (b) Let (Iu, Cu) and (Iv, Cv) denote a split partition of G[NG(u)⧵NG(v)] and G[NG(v)⧵NG(u)] respectively. For each pair of split partitions (Iu, Cu), (Iv, Cv) :

� (i) Construct solution candidate S ∶={u, v}∪(NG(u)∩NG(v)) ∪ Cu ∪ Cv (ii) If G ⊕ S is a member of G , return S

2. For each edge uv ∈ E(G) :

(a) If G[NG[u]∩NG[v]] is not a split graph, skip this iteration and try the next edge. (b) For each possible split partition (I, C) of G[NG[u]∩NG[v]] :

� (i) Construct solution candidate S ∶={u, v}∪C (ii) If G ⊕ S is a member of G , return S

3. Return ‘None’

Theorem 6 Let G be a class of triangle-free graphs such that deciding whether an G n-vertex graph is in is solvable in f(n) time for some function f. Then Subgraph G O 6 4 ⋅ Complement to is solvable in (n + n f (n)) time.

Proof We will prove that Algorithm 5 is correct, and that its running time is 4 2 O(n ⋅ (n + f (n))) . We begin by proving correctness. We have three cases to consider: (i) There exists a solution which hits a non-edge, (ii) there is a solution S such that it forms a clique of size at least two in G, and (iii) no solution exists. In the case that there exists a solution S hitting a non-edge uv, we will at some point guess this non-edge in Step 1 of the algorithm. By Lemma 3, we have (by symmetry) that both G[NG(u)⧵NG(v)] and G[NG(v)⧵NG(u)] are split graphs, so we do not miss the solution S in Step 1a. Since we try every possible combination of split partitions in Step 1b, we will by Lemma 3 at some point construct S correctly such � that S = S. In the case that there exist only solutions which hit exactly a clique, we frst fnd some edge uv ∈ E(G) . At some point we guess uv such that both endpoints are in the same solution S. By Lemma 4b we know that G[NG(u)∩NG(v)] is a split graph, so Algorithmica we will not miss S in Step 2a. Since we try every split partition in Step 2b, we will � by Lemma 4 at some point construct S correctly such that S = S. Lastly, in the case that there is no solution, we know that there neither exists a solution which hits a non-edge, nor a solution which hits a clique. Since these three cases exhaust the possibilities, we can correctly report that there is no solution when none was found in the previous steps. For the runtime, recall that determining whether a graph is split can be done in O(n) time given the degree sequence [12], and Lemma 2 establish that enumerating all split partitions can also be done in linear time. 2 The sub-procedure of Step 1 is performed at most O(n ) times, where Step 1a 2 2 takes O(n) time and Step 1b takes at most O(n ⋅ (n + f (n)) time. In total, Step 4 2 1 will take no longer than O(n ⋅ (n + f (n))) time. The sub-procedure of Step 2 2 is performed at most O(n ) times. Step 2a is done in O(n) time, and step 2b is 2 done in O(n ⋅ (n + f (n)) time; hence the asymptotic runtime of the entire step 2 is 3 2 O(n ⋅ (n + f (n))) . The worst running time among these steps is Step 1, and as such 4 2 the runtime of Algorithm 5 is O(n ⋅ (n + f (n))) . ◻

4 Complement to d‑Degenerate Graphs

For a constant integer d > 0 , we say that a graph G is d-degenerate, if every (not necessarily proper) subgraph of G has a vertex of degree at most d. For example, trees are 1-degenerate, while planar graphs are 5-degenerate.

Theorem 7 Let G be a class of d-degenerate graphs such that deciding whether an G n-vertex graph is in is solvable in f(n) time for some function f. Then Subgraph G ⋅ O∗(4d) Complement to is solvable in f (n) n time.

Proof Let G be an n-vertex graph. We are looking for a vertex subset S of G such that G ⊕ S ∈ G. We start from trying all vertex subsets of G of size at most 2d as a candidate for 2d S. Thus, in O(n ⋅ f (n)) time we either fnd a solution or conclude that a solution, if it exists, should be of size more than 2d. Now we assume that S > 2d . We try all subsets of V(G) of size 2d + 1 . Then if G can be complemented to G , at least one of these sets, say X, is a subset of S. In n total, we enumerate 2d + 1 sets. For each guess of X we consider the set Y of all vertices in V(G)⧵X with at least d + 1 neighbors in X. The observation here is that most vertices from Y are in S. For every size d + 1 subset X ⊆ X , there can be at most d vertices of Y⧵S who have X contained in their neighborhood—otherwise there remains a Kd+1,d+1 not killed by S. Because every vertex of Y has at least d + 1 neighbors in X, it follows that if more than X 2d + 1 = ⋅ d = ⋅ d d + 1 d + 1 Algorithmica vertices of Y are not in S, then G ⊕ S contains a complete bipartite graph Kd+1,d+1 as n G ⊕ S d a subgraph, and hence is not -degenerate. Thus, we make at most guesses on which subset of Y is (not) in S. Similarly, let the set Z contain all vertices from V(G)⧵X with at most d neighbors in X—we observe that most vertices of Z must be outside S. Analogous to before, for any particular size d + 1 subset X ⊆ X , at most d vertices of Z ∩ S can avoid X entirely in their neighborhood. Since every vertex in Z have at least d + 1 non-neighbors in X, it follows that at most of vertices from Z could belong to S. We make at n Z S V(G)=X ∪ Y ∪ Z most guesses of which vertices of are in . Since , if there is a solution S, it will be found in one from at most 2 n n ∗ d ⋅ = nO (4 ) 2d + 1 guesses. Since for each set S we can check in f(n) time whether G ⊕ S ∈ G , this concludes the proof. ◻

5 Complement to M‑partition

Many graph classes can be defned by whether it is possible to partition the vertices of graphs in the class such that certain constraints at met. For instance, a complete bipartite graph is one which can be partitioned into two vertex sets such that every edge between the two sets is present, and no edge exists within any of the partitions. Other graph classes which can be described in a similar manner are split graphs and k-colorable graphs. Feder et al. [8] formalized such partition properties of graph classes by making use of a symmetric matrix over {0, 1, ⋆} , called an M-partition.

Defnition 8 (M-partition) For a symmetric k × k matrix M over {0, 1, ⋆} , we say G that a graph G belongs to the graph class M if its vertices can be partitioned into k (possibly empty) sets X1, X2, … , Xk such that, for every i ∈[k]:

• if M[i, i]=1 , then Xi is a clique, and • if M[i, i]=0 , then Xi is an independent set, and for every i, j ∈[k] , i ≠ j,

• if M[i, j]=1 , then every vertex of Xi is adjacent to all vertices of Xj , and • if M[i, j]=0 , then there is no edge between Xi and Xj.

Note that if M[i, j]=⋆ , then there is no restriction on the edges between vertices from Xi and Xj . For example, for matrix Algorithmica

0 ⋆ M = ⋆ 0 the corresponding class of graphs is the class of bipartite graphs, while matrix 0 ⋆ M = ⋆ 1 identifes the class of split graphs. In this section we prove the following theorem. G G Theorem 9 Let = M be a graph class described by an M-partition matrix of size G 2 × 2 . Then Subgraph Complement to is solvable in polynomial time.

In particular, Theorem 9 yields polynomial-time algorithms for Subgraph Com- G G plement to when is the class of split graphs or (complete) bipartite graphs. The proof of our theorem is based on the following beautiful dichotomy result of G Feder et al. [8] on the recognition of classes M described by 4 × 4 matrices. Note that a symmetric k × k matrix M is said to contain another symmetric matrix M if there exists a set of indices X ⊆ [k] such that removing both rows an columns corresponding to X from M yields M .

Proposition 10 [8, Corollary 6.3] Suppose M is a symmetric 4 × 4 matrix over G {0, 1, ⋆} . Then the recognition problem for M is

• NP-complete when M contains the matrix for 3-coloring or its complement, and no diagonal entry is ⋆. • Polynomial time solvable otherwise.

Lemma 11 Let M be a symmetric k × k matrix giving rise to the graph class G G M = . Then there exists a symmetric 2k × 2k matrix M such that for any input G G G to Subgraph Complement to , it is a yes-instance if and only if G belongs to M . Moreover, M can be constructed in linear time such that there are equally many 1’s as 0’s on its diagonal.

Proof Given M, we construct a matrix M in linear time. We let M be a matrix of 2k × 2k M�[i, j] M[ i , j ] dimension , where entry is defned as 2 2 if at least one of ¬M[ i+1 , j+1 ] ¬1 = 0 ¬0 = 1 ¬⋆ = ⋆ i, j is even, and 2 2 if i, j are both odd. Here, , , and . For example, for matrix ⌈ ⌉ ⌈ ⌉ 0 ⋆ M = ⋆ 1 the above construction results in Algorithmica

10⋆⋆ 00⋆⋆ M� = . ⎛ ⋆⋆01⎞ ⎜ ⋆⋆11⎟ ⎜ ⎟ ⎜ ⎟ Observe that for each non-⋆ diagonal entry⎝ in M, two⎠ diagonal entries are created in M : one 1 and one 0. G It remains to show that an instance G to Subgraph Complement to is a yes- G instance if and only if G belongs to M . We prove the two directions of the claim separately. G G (⇒) Assume there is a subgraph complementation G ⊕ S into = M . Let � � � X1, X2, … , Xk be an M-partition of G ⊕ S . We defne partition X1, X2, … , X2k of G ≤ ≤ � as follows: for every vertex v ∈ Xi , 1 i k , we assign v to X2i−1 if v ∈ S and to X2i otherwise. We now show that every edge of G respects the requirements of M . Let uv ∈ E(G) be an edge, and let u ∈ Xi and v ∈ Xj . If at least one vertex from {u, v} , say v, is not in S, then uv is also an edge in G ⊕ S , thus M[i, j] ≠ 0 . Since v ∉ S , it � belongs to set v ∈ X2j . Vertex u is assigned to set X , where is either 2i or 2i − 1 , depending whether u belongs to S or not. But because 2j is even irrespectively of , � M [ ,2j]=M[i, j] ≠ 0. Now consider the case when both u, v ∈ S . Then the edge does not persist after the subgraph complementation by S, and thus M[i, j] ≠ 1 . We further know that u is � � assigned to X2i−1 and v to X2j−1 . Both 2i − 1 and 2j − 1 are odd, and by the construc- � tion of M , we have that M [2i − 1, 2j − 1] ≠ 0 , and again the edge uv respects M . An analogous argument shows that also all non-edges respect M . � � � (⇐) Assume that there is a partition X1, X2, … , X2k of G according to M . Let the set S consist of all vertices in odd-indexed parts of the partition. We now show that G ⊕ S can be partitioned according to M. We defne partition X1, X2, … , Xk by � assigning each vertex u ∈ X to X i . It remains to show that X1, X2, … , Xk is an i 2 M-partition of G ⊕ S. ⌈ ⌉ Let u ∈ Xi , v ∈ Xj . Suppose frst that uv ∈ E(G ⊕ S) . If at least one of u, v is not � in S, we assume without loss of generality that v ∉ S . Then uv ∈ E(G) and v ∈ X2j . � For vertex u ∈ X , irrespectively, whether is 2i or 2i − 1 , we have that � M [ ,2j]=M[i, j] ≠ 0 . Otherwise we have u, v ∈ S . Then uv is a non-edge in G, � and thus M [2i − 1, 2j − 1] ≠ 1 . But by the construction of M , we have that M[i, j] ≠ 0 , and there is no violation of M. An analogous argument shows that if u and v are not adjacent in G ⊕ S , it holds that M[i, j] ≠ 1 . Thus X1, X2, … , Xk is an M-partition of G ⊕ S , which concludes the proof. ◻

Now we are ready to prove Theorem 9.

Proof of Theorem 9 For a given 2 × 2 matrix M, we use Lemma 11 to construct a matrix M . Since M has equally many 1’s as 0’s on its diagonal, it can not contain the 3-coloring matrix. Then by Proposition 10, the recognition of whether G admits Algorithmica

G M -partition is in P. Thus, Subgraph Complement to is solvable in polynomial time. ◻

6 Subgraph Complementation to Graph Classes of Bounded Clique‑Width

G We show that Subgraph Complement to can be solved in polynomial time when G has bounded clique-width and can be expressed by an 1 property. We refer to the book [3] for the basic defnitions. We will use the following result of Hliněný and Oum [13].

Proposition 12 [13] There is an algorithm that for every integer k and graph G in 3 k+1 time O( V(G) ) either computes a (2 − 1) expression for a graph G or correctly concludes that the clique-width of G is more than k. Note that the algorithm of Hliněný and Oum only approximates the clique-width but does not provide an algorithm to construct an optimal k-expression tree for a graph G of clique-width at most k. But this approximation is usually sufcient for algorithmic purposes. Courcelle et al. [4] proved that every graph property that can be expressed in 1 can be recognized in linear time for graphs of bounded clique-width when given a k-expression.

Proposition 13 [4, Theorem 4] Let G be a class of graphs with clique-width bounded by constant k such that for each graph G ∈ G , a corresponding k-expres- O G sion can be found in (f ( V(G) , E(G) )) time. Then every 1 property on can be recognized in O(f ( V(G) , E(G) ) ⋅ g( , k))) time, for some function g. The nice property of graphs with bounded clique-width is that their subgraph complementation has also bounded clique-width. In particular, Kamiński, Lozin, and Milanič [17] observed that if G is a graph of clique-width k, then any subgraph complementation of G is of clique-width at most g(k) for some computable function g. For completeness, we here provide a more accurate upper bound.

Lemma 14 Let G be a graph, S ⊆ V(G) . Then CWD(G ⊕ S) ≤ 3CWD(G).

Proof Let CWD(G)=k . To show the bound, it is more convenient to use expression trees instead of k-expressions. An expression tree of a graph G is a rooted tree T with nodes of four types i, ∪̇ , and such that:

• Introduce nodes i(v) are leaves of T. For i ∈[k] and a vertex v ∈ V(G) , i(v) is associated to the singleton graph where vertex v is labeled by i. • Union node ∪̇ stands for a disjoint union of k-graphs associated with its children. Algorithmica

• Relabel node i→j has one child and is associated with the k-graph obtained by applying of the relabeling operation to the graph corresponding to its child. • Join node i,j has one child and is associated with the k-graph resulting by applying the join operation on the graph represented by its child: every vertex labeled with i is made adjacent to every vertex labeled by j. • The graph G is isomorphic to the graph associated with the root of T (with all labels removed).

The width of the tree T is the number of diferent labels appearing in T. If G is of clique-width k, then by parsing the corresponding k-expression, one can construct an expression tree of width k and, vise versa, given an expression tree of width k, it is straightforward to construct a k-expression. Throughout the proof we call the ele- ments of V(T) nodes to distinguish them from the vertices of G. Given a node x of an expression tree, Tx denotes the subtree of T rooted in x and the graph Gx represents the k-graph formed by Tx. An expression tree T is irredundant if for any join node i,j , the vertices labeled by i and j are not adjacent in the graph associated with its child. It was shown by Courcelle and Olariu [5] that every expression tree T of G can be transformed into an irredundant expression tree T of the same width in time linear in the size of T. Let T be an irredundant expression tree of G of width k rooted in r. We construct � the expression tree T for G = G ⊕ S by modifying T. Recall that the vertices of the graphs Gx for x ∈ V(T) are labeled 1, … , k . We introduce three groups of distinct labels 1, … , k , 1, … , k and 1, … , k . The labels 1, … , k and 1, … , k correspond the the labels 1, … , k for the vertices in S and V(G)⧵S respectively. The labels 1, … , k are auxiliary. Then for every node x of T we construct Tx using Tx starting the process from the leaves. We denote by Gx the k-graph corresponding to the root x of Tx. For every introduce node i(v), we construct an introduce node i(v) if v ∈ S and an introduce node i(v) if v ∉ S . Let x be a non-leaf node of T and assume that we already constructed the modifed expression trees of the children of x. Let x be a union node ∪̇ of T and let y and z be its children. i ∈{1, … , k} We construct k relabel nodes i,i for that form a path, make one end-node of the path adjacent to y in Ty and make the other end-node denoted by y the root of Ty constructed from Ty . Notice that in the corresponding graph Gy all the vertices of S are now labeled by 1, … , k instead of 1, … , k. (1) Next, we construct a union node ∪̇ denoted by x with the children y and z. This way we construct the disjoint union of Gy and Gz. Notice that vertices which are labeled by the same label in Gy and Gz are not adjacent in G. In other words, we should make the vertices of V(Gx)∩S and V(Gy)∩S G with the same label adjacent in . We achieve it by adding k join nodes i,i for i ∈{1, … , k} , forming a path out of them and making one end-node of the path (1) (2) adjacent to x . We declare the other end-node of the path denoted by x the new root. Observe now that for the set of vertices Yi of Gy labeled i and the set of vertices Zj of Gz labeled by j where i, j ∈{1, … , k} are distinct, it holds that the vertices of Yi Algorithmica and Zj are either pairwise adjacent in G or pairwise nonadjacent. More precisely, on this stage of construction we ensure that if the vertices of Yi are not adjacent to the vertices of Zj , then the vertices of Yi ∩ S and Zj ∩ S are made adjacent in G . To do it, for every two distinct i, j ∈{1, … , k} such that the vertices of Yi and Zj are not adjacent in G, construct a new join node i,j and form a path with all these nodes whose (2) (3) one end-node is adjacent to x and the other end-node x is the new root (we (3) (2) assume that x = x if have no new constructed nodes). i ∈{1, … , k} Finally, we add k relabel nodes i,i for that form a path, make one (3) end-node of the path adjacent to x and make the other end-node denoted by x the root of the obtained Tx . Clearly, all the vertices of S in Gx are labeled by 1, … , k. Let x be a relabel node i→j of T and let y be its child. We construct two relabel x x nodes i→j and i→j denoted by x and respectively. We make the child of x and we make the root y of Ty the child of x . Now, let x be a join node i→j of T and let y be its child. Recall that T is irredundant, that is, the vertices labeled by i and j in Gy are not adjacent. Clearly, we should avoid making adjacent the vertices in S in the construction of G . We do it by con- x, x , x structing three new join nodes i→j , j→i and i→j denoted by respectively. We make x the child of x, x the child of x and the node y of Ty is made the child of x . This completes the description of the construction of T . Using standard inductive arguments, it is straightforward to verify that G is isomorphic to the graph associ- � ated with the root of T , that is, CWD(G ) ≤ 3k . ◻ G Lemma 15 Let be an 1 property describing the graph class . Then there G(1) O exists an 1 property describing the graph class of size ∈ ( ). Proof We will construct from in the following way: We start by prepending ∃S ⊆ V(G) . Then for each assessment of the existence of an edge in , say uv ∈ E(G) , replace that term with ((u ∉ S ∨ v ∉ S)∧uv ∈ E(G))∨(u ∈ S ∧ v ∈ S ∧ uv ∉ E(G)) . Symmetrically, for each assessment of the non-existence of an edge uv ∉ E(G) , replace that term with ((u ∉ S ∨ v ∉ S)∧uv ∉ E(G))∨(u ∈ S ∧ v ∈ S ∧ uv ∈ E(G)). We observe that if is satisfable for some graph G, then for every S ⊆ V(G) , the subgraph complementation G ⊕ S will yield a satisfying assignment to . Con- versely, if is satisfable for a graph G, then there exist some S such that is satis- fed for G ⊕ S . For the size, we note that each existence check for edges blows up by a constant factor. ◻

We are ready to prove the main result of this section. G Theorem 16 Let be a graph class expressible in 1 which has clique-width G bounded by constant k. Then Subgraph Complement to is solvable in polynomial time. G Proof Let be the 1 formula which describes , and let G be an n-vertex input 3 � 3k+1 graph. We apply Proposition 12 for G and in O(n ) time either obtain a k ≤ 2 − 1 Algorithmica

-expression for G or conclude that the clique-width of G is more than 3k. In the lat- ter case, by Lemma 14, G cannot be subgraph complemented to G. We then obtain an 1 formula from Lemma 15, and apply Proposition 13, 3 � which works in O(n ⋅ g( , k )) time for some function g. As and k depend only on 3 constants and k, the runtime of the algorithm is O(n ) . ◻

We remark that if clique-width expression is provided along with the input graphs, G and can be expressed in 1 , then there is a linear time algorithm for Subgraph G Complement to . This follows directly from Lemma 15 and Proposition 13. Theorem 16 implies that for every class of graphs G of bounded clique-width characterized by a fnite set of fnite forbidden induced subgraphs, e.g. P4-free graphs (also known as cographs) or classes of graphs discussed in [1], the Subgraph G Complement to problem is solvable in polynomial time. However, Theorem 16 G does not imply that Subgraph Complement to is solvable in polynomial time for G being of the class of graphs having clique-width at most k. This is because such a G G class cannot be described by 1 . Interestingly, for the related class of graphs of bounded rank-width (see [5] for the defnition) at most k, the result of Oum and G Courcelle [6] combined with Theorem 16 implies that Subgraph Complement to is solvable in polynomial time.

7 Hardness of Subgraph Complementation to Regular Graphs

Let us remind that a graph G is regular if all its vertices have the same degree. We G G consider Subgraph Complement to where is the class of regular graphs, which we call Subgraph Complement to Regular Graph (SCR). In this section, we show that this problem is NP-complete by a reduction from Clique in Regular Graph.

We will need the following well-known proposition.

Proposition 17 [10] Clique in Regular Graph is NP-complete.

We remark that the above proposition follows because the Independent Set problem is NP-complete for cubic graphs [10]; fnding an independent set is equivalent to fnding a clique in the complement graph, and the complement of a regular graph is also regular. We are now ready to prove that Clique in Regular Graph is many-one reducible to Subgraph Complement to Regular Graph . We begin by defning a gadget which we will use in the reduction. Algorithmica

Defnition 18 (Gadget Hk,r ) For integers k ≥ 2 and r ≥ k such that r − k is even, we build the gadget graph Hk,r as follows. Initially, we let Hk,r consist of one clique on k − 1 vertices, as well as k − 1 distinct copies of Kr,r . These are all the vertices of the gadget, which is a total of (k − 1)+2r ⋅ (k − 1) vertices. We denote the vertices of the clique c1, c2, … , ck−1 , and we let the complete bipartite graphs be denoted 1 2 k−1 i by Kr,r, Kr,r, … , Kr,r . For a bipartite graph Kr,r , let the vertices of the two parts be i i i i i i denoted by a1, a2, … , ar and b1, b2, … , br respectively. If r = k , then the construction is already complete. Otherwise, we will now do some rewiring of the edges to complete the construction of Hk,r . Recall that r − k is i i i even. For each vertex ci of the clique, add one edge from ci to each of a1, a2, … , a r−k . 2 i i i Similarly, add an edge from ci to each of b1, b2, … , b r−k . Now remove the edges 2 i i i i i i a1b1, a2b2, … , a r−k b r−k . Once this is done for every i ∈[k − 1] , the construction is 2 2 complete. See Fig. 1.

i i We observe the following property of vertices aj , bj , and ci of Hk,r.

i Observation 19 For every i ∈[k − 1] and j ∈[r] , it holds that the degrees of aj i and bj in Hk,r are both exactly r, whereas the degree of ci is r − 2.

We now present our reduction in the form of an algorithm.

Algorithm 20 (Reduction KR to SCR) Input: An instance (G, k) of KR. Let r denote the regularity of G (the degree of its vertices). Output: An instance G of SCR such that it is a yes-instance if and only if (G, k) is a yes-instance of KR.

1. If k < 7 or k > r , solve the instance of KR by brute force. If it is a yes-instance, return a trivial yes-instance to SCR, if it is a no-instance, return a trivial no- instance to SCR. 2. If r − k is odd, modify G by taking two copies of G which are joined by a perfect matching between corresponding vertices. Then r increases by one, whereas k remains the same. Going forth, we assume G (and r) have already undergone this transformation if required. 3. Construct the graph G by taking the disjoint union of G and the gadget Hk,r . Return G .

H Fig. 1 The gadget graph k,r is built of k parts, namely a clique K k 1 k−1 , and − complete bipar- K1 , , Kk−1 tite graphs r,r … r,r with some rewiring Algorithmica

Let n = V(G) . We observe that the number of vertices in the returned instance 2 is at most 2n +(k − 1)+2r ⋅ (k − 1) , which is O(n ) . The running time of the algo- 7 rithm is O(n ) and thus is polynomial. Correctness of the reduction follows from the following two lemmata.

Lemma 21 Let (G, k) be the input of Algorithm 20, and let G be the returned result. If (G, k) is a yes-instance to Clique in Regular Graph, then G is a yes-instance of Subgraph Complement to Regular Graph .

Proof Let C ⊆ V(G) be a clique of size k in G. If the clique is found in step 1, then G is a trivial yes-instance, so the claim holds. Thus, we can assume that the graph G was constructed in step 3. If G was altered in step 2, we let C be the clique in one � of the two copies that was created. Let S ⊆ V(G ) consist of the vertices of C as well as the vertices of the clique Kk−1 of the gadget Hk,r . We claim that S is a valid solution to G . We show that G ⊕ S is r-regular. Any vertex not in S will have the same number of neighbors as it had in G . Since the only vertices that weren’t originally of degree r were those in the gadget clique Kk−1 , all vertices outside S also have degree r in G ⊕ S . What remains is to examine the degrees of vertices of C and of Kk−1. Let ci be a vertex of Kk−1 in G . Then ci lost its k − 2 neighbors from Kk−1 , gained i k neighbors from C, and kept r − k neighbors in Kr,r . We see that its new neighborhood has size k + r − k = r. Let u ∈ C be a vertex of the clique from G. Then u lost k − 1 neighbors from C, gained k − 1 neighbors from Kk−1 , and kept r −(k − 1) neighbors from G − C . In total, u will have r −(k − 1)+(k − 1)=r neighbors in G ⊕ S . Since every vertex of G ⊕ S has degree r, it is r-regular, and thus G is a yes-instance. ◻

Lemma 22 Let (G, k) be the input of Algorithm 20, and let G be the returned result. If G is a yes-instance to Subgraph Complement to Regular Graph , then (G, k) is a yes-instance of Clique in Regular Graph.

� Proof Let S ⊆ V(G ) be a solution witnessing that G is a yes-instance. If G was the trivial yes-instance returned in step 1 of Algorithm 20, the statement trivially holds. Going forward we may thus assume G was returned in step 3, and that k ≥ 7. Our frst claim is that G ⊕ S is regular with regularity r. Assume for the sake of contradiction it is regular with regularity r ≠ r . Note that if r is any other value than r − 2 , then every vertex of G must be in S, as degrees of vertices outside S do not change under subgraph complementation; but then G ⊕ S is simply the complement of G , and it is not regular. So r must be r − 2 . Then every vertex except those in the gadget clique Kk−1 must be in S. Consider a vertex in one of the complete bipartite graphs from the gadget—it will have at least n new incident edges (from the graph G). Since n > r , the vertex will not have r − 2 neighbors in G ⊕ S , a contradiction. We have established that G ⊕ S is r-regular. It must thus be the case that every vertex of the gadget clique Kk−1 is in S, since these vertices do not have degree r in G . Algorithmica

Our second claim is that S = 2k − 1 , and moreover, that no neighbor of Kk−1 is from S in G . To show this, we let p = S⧵Kk−1 , and proceed to show that p = k . Towards this end, consider a vertex ci ∈ Kk−1 . This vertex has some number of neighbors in S⧵Kk−1 , denoted xi = NG� (ci)∩(S⧵Kk−1) . We know that ci has r neighbors in G ⊕ S . Let us count them: some neighbors are preserved by i the subgraph complementation, namely r − k − xi of its neighbors, found in Kr,r . Some neighbors are gained, namely p − xi of the vertices in S. Thus, we have that r = r − k − x + p − x x = p−k i i . The r’s cancel, and we get i 2 . This is true for every i ∈[k − 1] , so we simply denote the number by x = xi , and get p = k + 2x. Towards the claim, it remains to show that x = 0 . Because the neighborhoods of ⋅ distinct ci and cj are disjoint outside Kk−1 in G , we get that p ≥ (k − 1) x . We sub- stitute p, and get k + 2x ≥(k − 1) ⋅ x k ≥(k − 3) ⋅ x k ≥x k − 3

Recalling that k ≥ 7 , we have that x is either 1 or 0. Assume for the sake of contra- i diction that x = 1 . Then without loss of generality, each ci has some neighbor aj i i which is in S. Since aj had degree r in G , it must hold that aj has equally many i i neighbors as non-neighbors in S. At most one of aj ’s neighbors in S is from Kr,r in G S −3 Ki k ≥ 7 Ki , this means that at least 2 vertices of r,r are in S. Because and the r,r ’s are completely disjoint in G for diferent values of i ∈[k − 1] , we get that S − 3 S − 3 S ≥ ⋅ (k − 1) ≥ ⋅ 6 2 2 S ≥3 ⋅ S − 9 9 ≥2 ⋅ S Seeing that S ≥ k − 1 ≥ 6 , this is a contradiction. Thus, x must be 0, so p = k + 2x = k and the second claim holds. We now show that S⧵Kk−1 is a clique in G . Assume for the sake of contradiction � it is not, and let u, v ∈ S⧵Kk−1 be vertices such that uv ∉ E(G ) . Consider the vertex u. By the second claim above we know that u does not have a neighbor from Kk−1 in G . It will thus gain at least k edges going to Kk−1 ∪{v} , and lose at most k − 2 edges going to S⧵(Kk−1 ∪{u, v}) . Because u was of degree r in G yet gained more edges than it lost by the subgraph complementation, its degree is strictly greater than r in G ⊕ S . This is a contradiction, hence S⧵Kk−1 is a clique in G . Because k ≥ 3 , the clique S⧵Kk−1 can not be contained in the gadget Hk,r nor span across both copies of G created in step 2 of the reduction (if that step was applied). It must therefore be contained in the original G. Thus, G has a clique of size k, and (G, k) is a yes-instance of Clique in Regular Graph. ◻ Algorithmica

Theorem 23 Subgraph Complement to Regular Graphis NP-complete.

Proof Lemmata 21 and 22 together with Proposition 17 conclude the proof of NP- hardness. Membership in NP is trivial, so NP-completeness holds. ◻

We remark that if the regularity r is a constant, the problem becomes polynomial time solvable by Theorem 7.

8 Conclusion and Open Problems

G In this paper we initiated the study of Subgraph Complement to . Many interesting questions remain open. In particular, what is the complexity of the problem when G is

• the class of chordal graphs, • the class of interval graphs, • the class of graph excluding a path P5 as an induced subgraph, • the class of graphs with min degree ≥ r for some constant r.

(1) Another natural question is to consider a class C , such that C can be recognized in C C polynomial time. Let ( 1denote) the class of complement graphs to . Is it then also possible to recognize C in polynomial time? More broadly, it is also interesting to see what happens as we allow more than (k) one subgraph complementation; how quickly can we recognize the class G for some class G ? It will also be interesting to investigate what happens if we com- bine subgraph complementation with other graph modifcations, such as the Seidel switch.

Acknowledgements Open Access funding provided by University of Bergen. We thank Saket Saurabh for helpful discussions, and also a great thanks to the anonymous reviewers who provided valuable feed- back. The frst three authors have been supported by the Research Council of Norway via the projects “CLASSIS” and “MULTIVAL”. The fourth author has been supported by project “DEMOGRAPH” (ANR-16-CE40-0028). An extended abstract of this paper was presented in [9].

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